转换为极坐标形式令x=ρcosθ y=ρsinθx²+y²≤x+y 所以 ρ≤cosθ+sinθ∫∫▁D(x+y)dσ=∫[-π/4,3π/4] dθ∫[0,cosθ+sinθ] ρ(cosθ+sinθ) ρdρ=(1/3)∫[-π/4,3π/4] (cosθ+sinθ)^4 dθ=(1/3)...